Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define nn 65536*8
ll a[nn];ll sa[nn];
ll b[nn];ll n;ll mod;
map<ll,ll> m;
pair<ll,ll> range[nn];
ll rf[nn];
ll modi(ll x)
{
	return (x%mod+mod)%mod;
}
int tot=0;
struct travellerspride{
	ll tree[nn*2+10];
	int llc[nn*2+10],rrc[nn*2+10];
	ll tag[nn];ll nul=-1e17;int all;
	#define lc llc[id]
	#define rc rrc[id]
	void up(int id)
	{
		tree[id]=min(tree[lc],tree[rc]);
	}
	void set(int id,ll y)
	{
		tree[id]=y;
		tag[id]=y;
	}
	void down(int id)
	{
		if(tag[id]>nul)
		{
			set(lc,tag[id]);
			set(rc,tag[id]);
		}
		tag[id]=nul;
	}
	#define mid (l+r)/2
	void build(int l,int r,int &id)
	{
		if(!id) id=++all;
		tag[id]=nul;
		if(l==r)
		{
			if(l>tot) tree[id]=0-rf[l-tot];
			else tree[id]=0-rf[l];
			return ;
		}
		build(l,mid,lc);build(mid+1,r,rc);
		up(id);
	}
	ll query(int a,int b,int l,int r,int id)
	{
		if(l>=a and r<=b)
			return tree[id];
		down(id);
		if(b<=mid) return query(a,b,l,mid,lc);
		else if(a>mid) return query(a,b,mid+1,r,rc);
		else return min(query(a,b,l,mid,lc),query(a,b,mid+1,r,rc));
	}
	void update(int a,int b,int l,int r,int id,ll y)
	{
		if(a>b) return ;
		if(l>=a and r<=b)
		{
			set(id,y);
			return ;
		}
		down(id);
		if(a<=mid) update(a,b,l,mid,lc,y);
		if(b>mid) update(a,b,mid+1,r,rc,y);
		up(id);
	}
	ll query(int l,int r,int id)
	{
		if(l==r)
		{
			if(l>tot) return tree[id]+rf[l-tot];
			else return tree[id]+rf[l];
		}
		down(id);
		return min(query(l,mid,lc),query(mid+1,r,rc));
	}
	
}trcyx;


int main()
{
	scanf("%lld%lld",&n,&mod);
	set<ll> s;
	for(int i=0;i<n;i++) 
	{
		scanf("%lld%lld",&a[i],&b[i]);
		if(b[i]==1 and a[i]*2>mod)
		{
			puts("-1");
			exit(0);
		}
		if(i)a[i]+=a[i-1];//aishouldnotbe modded
		if(2==b[i]) continue;
		ll r=-2*a[i],l=0;if(i) l=-2*a[i-1];//this range is what we can stay
		l=modi(l),r=modi(r);
		range[i]=make_pair(l,r);
		s.insert(l);s.insert(r);
	}
	for(set<ll>::iterator it=s.begin();it!=s.end();it++)
		m[*it]=++tot,rf[tot]=*it;
	int rt=0;ll totot=tot+tot;
	trcyx.build(1,totot,rt);
	for(int i=0;i<n;i++)
	{
		ll l=range[i].first,r=range[i].second;
		l=m[l],r=m[r];if(l>r) r+=tot;//thus we can have real ranges
		if(b[i]==1)
		{
			ll fl=trcyx.query(1,l,1,totot,rt)+rf[l];
			fl=min(fl,trcyx.query(l,totot,1,totot,rt)+rf[l]+mod);
			trcyx.update(l,l,1,totot,rt,fl-rf[l]);
			trcyx.update(l+tot,l+tot,1,totot,rt,fl-rf[l]);
			//these poins are same
			l=range[i].first,r=range[i].second;
			l=m[l],r=m[r];
			if(l<=r)
			{
				trcyx.update(r+1,tot,1,totot,rt,1e17);
				trcyx.update(1,l-1,1,totot,rt,1e17);
				
				trcyx.update(r+1+tot,totot,1,totot,rt,1e17);
				trcyx.update(1+tot,l-1+tot,1,totot,rt,1e17);
			}
			else
			{
				trcyx.update(r+1,l-1,1,totot,rt,1e17);
				
				trcyx.update(r+1+tot,l-1+tot,1,totot,rt,1e17);
			}
			//thus these ranges are mpossible
		}
	}
	cout<<trcyx.query(1,totot,rt)+a[n-1]*2;
}